For all permissible values of `x`, if `y= (sin3x(cos6x+cos4x))/(sinx(cos8x+cos2x))`, then range of `y` is `(-oo,a)uu(b,oo)` If `2b` is the first term of a G.P and `a` is its common ratio then

Options

A. `b - a = 10/3`
B. `3a+b=4`
C. `S_(oo) = 9`
D. `S_(oo) = 27/10(a+b)`

Answer

To solve the given problem step by step, we will analyze the expression for \( y \) and find its range.

### Step 1: Simplify the Expression for \( y \)

Given:
\[ y = \frac{\sin 3x (\cos 6x + \cos 4x)}{\sin x (\cos 8x + \cos 2x)} \]

Using the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

We can simplify \( \cos 6x + \cos 4x \):
\[ \cos 6x + \cos 4x = 2 \cos\left(\frac{6x + 4x}{2}\right) \cos\left(\frac{6x - 4x}{2}\right) = 2 \cos(5x) \cos(x) \]

Similarly, for \( \cos 8x + \cos 2x \):
\[ \cos 8x + \cos 2x = 2 \cos\left(\frac{8x + 2x}{2}\right) \cos\left(\frac{8x - 2x}{2}\right) = 2 \cos(5x) \cos(3x) \]

Substituting these back into the expression for \( y \):
\[ y = \frac{\sin 3x \cdot 2 \cos(5x) \cos(x)}{\sin x \cdot 2 \cos(5x) \cos(3x)} \]

The \( 2 \cos(5x) \) cancels out (assuming \( \cos(5x) \neq 0 \)):
\[ y = \frac{\sin 3x \cdot \cos x}{\sin x \cdot \cos 3x} \]

### Step 2: Rewrite Using Tangent

We can rewrite the expression using tangent:
\[ y = \frac{\tan 3x}{\tan x} \]

### Step 3: Analyze the Range of \( y \)

Using the identity for \( \tan 3x \):
\[ \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \]

Let \( a = \tan x \):
\[ y = \frac{3a - a^3}{1 - 3a^2} \]

This can be rearranged into a quadratic form:
\[ y(1 - 3a^2) = 3a - a^3 \]
\[ a^3 - 3a + 3ya^2 - y = 0 \]

### Step 4: Find the Discriminant

To find the range of \( y \), we need the discriminant \( D \) of this cubic equation to be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Where:
- \( a = 1 \)
- \( b = 3y \)
- \( c = -3 \)

Thus:
\[ D = (3y)^2 - 4(1)(-y) = 9y^2 + 4y \geq 0 \]

### Step 5: Solve the Inequality

Factoring gives:
\[ y(9y + 4) \geq 0 \]

The critical points are:
\[ y = 0 \quad \text{and} \quad 9y + 4 = 0 \Rightarrow y = -\frac{4}{9} \]

### Step 6: Determine the Range

The intervals where the product is non-negative are:
1. \( y \leq -\frac{4}{9} \)
2. \( y \geq 0 \)

Thus, the range of \( y \) is:
\[ (-\infty, -\frac{4}{9}] \cup [0, \infty) \]

### Step 7: Identify \( a \) and \( b \)

From the range, we have:
- \( a = -\frac{4}{9} \)
- \( b = 0 \)

### Step 8: Find \( 2b \) and the Common Ratio

Given \( 2b \) is the first term of a G.P.:
\[ 2b = 0 \quad \text{(first term)} \]
And \( a \) is the common ratio:
\[ a = -\frac{4}{9} \]

### Conclusion

Thus, the values of \( a \) and \( b \) are:
- \( a = -\frac{4}{9} \)
- \( b = 0 \)